Going by the Doppler effect formula, when a source A with speed Vs is moving towards an observer which is also traveling towards the source A with the same speed of Vs from the opposite direction, does that mean the "Apparent" frequency is not defined?
1 Answer
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The formula for Doppler effect is $f^{'} = f\frac{v+v_{o}}{v+v_{s}}$,
Where $f^{'}$ is the apparent frequency, $f$ is the original frequency of the source, $v$ is the speed of sound, $v_{o}$ is the observer's speed and $v_{s}$ is the source's speed.
As such assuming that the observer moves towards the source with a velocity $v_{s}$ (hence the positive sign), and the source move towards the observer with a velocity $-v_{s}$ (hence the negative sign), then the formula transforms to: $f^{'} = f\frac{v+v_{s}}{v-v_{s}}$
Hence the Doppler's apparent frequency cannot be undefined.
answered Apr 10, 2018 at 20:51
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$\begingroup$ That's when it's moving away from the Observer $\endgroup$ Commented Apr 10, 2018 at 20:52
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$\begingroup$ No, the term for velocities for the observer is in the numerator, while that for the source is in the denominator. $\endgroup$ Commented Apr 10, 2018 at 20:58